poj 1163 The Triangle(dp)

The Triangle

Description

73   88   1   02   7   4   44   5   2   6   5

(Figure 1)

Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right.

Input

Your program is to read from standard input. The first line contains one integer N: the number of rows in the triangle. The following N lines describe the data of the triangle. The number of rows in the triangle is > 1 but <= 100. The numbers in the triangle, all integers, are between 0 and 99.

Output

Your program is to write to standard output. The highest sum is written as an integer.

Sample Input

5
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5

Sample Output

30

第一道DP题，水过，纪念一下~

Java AC 代码

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int rows = sc.nextInt();
        int input[][] = new int[rows + 1][rows + 1];
        int result[][] = new int[rows + 1][rows + 1]; //将每个点的最大结果存在数组里
        for(int i = 1; i <= rows; i++)
            for( int j = 1; j <= i; j++) {
                input[i][j] = sc.nextInt();
            }
        result[1][1] = input[1][1];
        dp(input, result);

        int max = Integer.MIN_VALUE;
        for(int i = 1; i <= rows; i++) {   //找出最后一行最大的一个，即为结果
            if(result[rows][i] > max)
                max = result[rows][i];
        }
        System.out.println(max);
    }

    public static void dp(int[][] input, int[][] result) {

        int rows = input.length - 1;
        for(int i = 2; i <= rows; i++)
            for(int j = 1; j <= i; j++) {
                if(j == 1)                      //每行的第一列的最大和 只能由上一行的第一列的最大和得到
                    result[i][j] = result[i - 1][j] + input[i][j];
                else if(j == i)            //每行的最后一列的最大和 只能由上一行的最后一列的最大和得到
                    result[i][j] = result[i - 1][j - 1] + input[i][j];
                else                 //其他的则是可以由两个方向中大的那个得到
                    result[i][j] = Math.max(result[i - 1][j - 1], result[i - 1][j]) + input[i][j];
            }
    }
}